By Proposition REF , we have \(\mid Z(\frac{G}{Z(G)})\mid =1\) . Again, since \(\frac{G}{Z(G)}\)  is metacyclic therefore \((\frac{G}{Z(G)})^{\prime }\)  is cyclic. We have \(\frac{G}{Z(G)}\)  has a normal subgroup \(\frac{N}{Z(G)}\)  of index \(p\)  and hence \((\frac{G}{Z(G)})^{\prime } \subseteq \frac{N}{Z(G)}\) . Consequently \(\mid (\frac{G}{Z(G)})^{\prime } \mid =q, r\)  or \(qr\) .
Now, if \(\mid (\frac{G}{Z(G)})^{\prime } \mid =q\) , then \(\frac{N}{Z(G)}\)  is a cyclic normal subgroup of \(\frac{G}{Z(G)}\)  of index \(p\) . But then by [1]}, we have \(\mid \frac{G}{Z(G)}\mid \ne 1\) , which is a contradiction.
Next suppose \(\mid (\frac{G}{Z(G)})^{\prime } \mid =r\) . Now, if \((\frac{G}{Z(G)})^{\prime } \subsetneq C(xZ(G))\)  for some \(xZ(G) \in (\frac{G}{Z(G)})^{\prime } \setminus \lbrace 1 \rbrace \) , then by Remark REF , \(C(xZ(G))\)  is an abelian normal subgroup of \(\frac{G}{Z(G)}\)  of prime index. Consequently, by [1]} we have \(\mid \frac{G}{Z(G)}\mid \ne 1\) , which is a contradiction
On the otherhand, suppose \((\frac{G}{Z(G)})^{\prime } =C(xZ(G))\)  for any \(xZ(G) \in (\frac{G}{Z(G)})^{\prime } \setminus \lbrace 1 \rbrace \) . Then by [3]}, \(\frac{G}{Z(G)}\)  is a Frobenious group with Frobenious kernel \(\frac{C(x)}{Z(G)}=C(xZ(G))=(\frac{G}{Z(G)})^{\prime }\)  and cyclic Frobenius complement \(\langle yZ(G)\rangle = \frac{C(y)}{Z(G)}\)  of order \(pq\) . Moreover, \(C(y)\)  is abelian by Remark REF . Therefore by [4]}, we have \(\mid \operatorname{Cent}(G) \mid = r+2\) .
Finally, If \(\mid (\frac{G}{Z(G)})^{\prime } \mid =qr\) , then \(\frac{G}{Z(G)} \cong C_{qr} {\rtimes }_\theta C_p\)  and hence by [5]}, \(\mid \operatorname{Cent}(G) \mid = qr+2\) .
